Here's the question:
- $\overset{\Delta}{ABC}$ is a triangle.
- $D$ is a point on $[BC]$.
- $|BD|=4$.
- $|AD|=|CD|$.
- $\text m(\widehat{CBA})=\alpha=30^\circ$.
- $\text m(\widehat{ACB})=\beta=20^\circ$.
- What is $\color{magenta}{|AC|}=\color{magenta}x$?
Geometric approaches gave me nothing. Letting $|AD|=|CD|=a$, $|AB|=b$ and applying law of cosines to all three triangles yields;
$$ \begin{align} x&=2a\cos 20^\circ\\ 16&=a^2+b^2+2ab\cos 70^\circ\\ (a+4)^2&=b^2+x^2+2bx\cos 50^\circ \end{align} $$
respectively, from $\overset{\Delta}{ACD}$, $\overset{\Delta}{ABD}$ and $\overset{\Delta}{ABC}$. Directly isolating $x$ from these equations gives an expression involves trigonometric function(s), and i cannot find a way to annihilate trigonometric functions in these equations to reach the answer which is $\color{magenta}x=\color{magenta}4$.